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%% productTopology.tex
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%% Made by Alex Nelson
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%% Started on  Tue Jun  2 19:41:14 2009 Alex Nelson
%% Last update Tue Jun  2 19:41:14 2009 Alex Nelson
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\begin{prob}
If we have two topological spaces $X$ and $Y$, can we ``glue''
them together? That is, given two topological spaces, we wish to
construct a new one using only what we know of $X$ and $Y$.
\end{prob}

\begin{defn}\label{defn:productTopology}
Let $X$, $Y$ be topological spaces. The \textbf{Product Topology}
on $X\times Y$ is the topology having as basis the collection
$\mathscr{B}$ of all sets of the form $U\times V$, where $U$ is
an open subset of $X$ and $V$ is an open subset of $Y$.
\end{defn}

\begin{rmk}\label{rmk:productTopologyBasis}
Being rigorous, we should probably verify that this $\mathscr{B}$
beast really is a basis. For any $x\times y\in X\times Y$, we see
that $X\times Y\in\mathscr{B}$ so the first condition is
trivially satisfied. The second condition, let $(x\times y)\in
U_{1}\times V_{1}$ and $(x\times y)\in U_{2}\times V_{2}$. We see
that $\mathscr{B}$ is the collection of the product of all open
subsets, which allows us to see that
\begin{equation}%\label{eq:}
(U_{1}\times V_{1})\cap(U_{2}\times V_{2}) = (U_{1}\cap
  U_{2})\times(V_{1}\cap V_{2})
\end{equation}
is also the product of open sets, so it's a basis element. Thus
the second property of a basis is satisfied.
\end{rmk}
\begin{thm}\label{thm:productOfBases}
If $\mathscr{B}$ is a basis for the topology of $X$ and
$\mathscr{C}$ is a basis for the topology of $Y$, then the
collection
\begin{equation}%\label{eq:}
\mathscr{D} = \{B\times C|B\in\mathscr{B},C\in\mathscr{C}\}
\end{equation}
is a basis for the topology of $X\times Y$.
\end{thm}
\begin{proof}
We will use lemma \ref{lem:findingTopologyBasis} to prove
this. That is, for some open set $W\subset X\times Y$ and each
$x\times y\in W$ there is a basis element $U\in\mathscr{B}$ such
that $x\in U$ and a basis element $V\in\mathscr{C}$ such that
$y\in V$, so $x\times y\in U\times V\subset W$ and $U\times
V\in\mathscr{D}$. Thus $\mathscr{D}$ is a basis.
\end{proof}
\begin{defn}\label{defn:projection}
Let $\pi_{1}:X\times Y\to X$ be defined by the equation
\begin{equation}%\label{eq:}
\pi_{1}(x,y)=x;
\end{equation}
let $\pi_{2}:X\times Y\to Y$ be defined by the equation
\begin{equation}%\label{eq:}
\pi_{2}(x,y) = y.
\end{equation}
The maps $\pi_{1}$ and $\pi_{2}$ are called the
\textbf{projections} of $X\times Y$ onto its first and second
factors, respectively.
\end{defn}
\begin{rmk}\label{rmk:projectionsAreSurjective}
Observe that projections are surjective provided that both $X$
and $Y$ are nonempty. If one is empty, $X\times Y$ is empty too,
and everything becomes trivial.
\end{rmk}
\begin{rmk}\label{rmk:inverseOfProjections}
Note that if $U\subset X$ is open, $\pi_{1}^{-1}(U)=U\times
Y$. Similarly, if $V\subset Y$ is open, $\pi_{2}^{-1}(V)=X\times
V$. Their intersection is $U\times V$.
\end{rmk}

\begin{thm}\label{thm:subBasisForProductTopology}
The collection
\begin{equation}%\label{eq:}
\mathscr{S}=\{\pi_{1}^{-1}(U)|U\text{ is open in
}X\}\cup\{\pi_{2}^{-1}(V)|V\text{ is open in }Y\}
\end{equation}
is a subbasis for the product topology on $X\times Y$.
\end{thm}
\begin{proof}
Let $\mathcal{T}$ be the topology on $X\times Y$, $\mathcal{T}'$
be the topology generated by our subbasis. We see that each
element $W\in\mathscr{S}$ is an open subset of the product
topology, which means that the topology generated by
$\mathscr{S}$ is contained in $\mathcal{T}$, i.e.
\begin{equation}%\label{eq:}
\mathcal{T}'\subset\mathcal{T}.
\end{equation}
We need to show that $\mathcal{T}\subset\mathcal{T}'$. We see
given a basis element $U\times V$ for $\mathcal{T}$ that
\begin{equation}%\label{eq:}
U\times V = \pi_{1}^{-1}(U)\cap\pi_{2}^{-1}(V)
\end{equation}
which is nothing more than a finite intersection of subbasis
elements. This implies that $\mathcal{T}\subset\mathcal{T}'$ as desired.
\end{proof}
